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91 lines
5.6 KiB
Markdown
91 lines
5.6 KiB
Markdown
# Binary search insertion
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Binary search is not only used to search for target elements but also to solve many variant problems, such as searching for the insertion position of target elements.
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## Case with no duplicate elements
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!!! question
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Given a sorted array `nums` of length $n$ with unique elements and an element `target`. Now insert `target` into `nums` while maintaining its sorted order. If `target` already exists in the array, insert it to the left of the existing element. Please return the index of `target` in the array after insertion. See the example shown in the figure below.
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If you want to reuse the binary search code from the previous section, you need to answer the following two questions.
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**Question one**: If the array already contains `target`, would the insertion point be the index of existing element?
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The requirement to insert `target` to the left of equal elements means that the newly inserted `target` will replace the original `target` position. In other words, **when the array contains `target`, the insertion point is indeed the index of that `target`**.
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**Question two**: When the array does not contain `target`, at which index would it be inserted?
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Let's further consider the binary search process: when `nums[m] < target`, pointer $i$ moves, meaning that pointer $i$ is approaching an element greater than or equal to `target`. Similarly, pointer $j$ is always approaching an element less than or equal to `target`.
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Therefore, at the end of the binary, it is certain that: $i$ points to the first element greater than `target`, and $j$ points to the first element less than `target`. **It is easy to see that when the array does not contain `target`, the insertion point is $i$**. The code is as follows:
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```src
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[file]{binary_search_insertion}-[class]{}-[func]{binary_search_insertion_simple}
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```
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## Case with duplicate elements
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!!! question
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Based on the previous question, assume the array may contain duplicate elements, all else remains the same.
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Suppose there are multiple `target`s in the array, a regular binary search can only return the index of one `target`, **and it cannot determine how many `target`s are to the left and right of that it**.
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The problem requires inserting the target element to the very left, **so we need to find the index of the leftmost `target` in the array**. Initially consider implementing this through the steps shown in the figure below.
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1. Perform a binary search to find any `target`'s index, say $k$.
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2. Starting from index $k$, perform a linear search to the left until the leftmost `target` is found and return.
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Although this method is feasible, it includes linear search, so its time complexity is $O(n)$. This method is inefficient when the array contains many duplicate `target`s.
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Now consider extending the binary search code. As shown in the figure below, the overall process remains the same. In each round, we first calculate the middle index $m$, then compare the value of `target` and `nums[m]`, which results in the following cases.
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- When `nums[m] < target` or `nums[m] > target`, it means `target` has not been found yet, thus use the normal binary search to narrow the search range, **bring the pointers $i$ and $j$ closer to `target`**.
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- When `nums[m] == target`, it indicates that the elements less than `target` are in the range $[i, m - 1]$, therefore use $j = m - 1$ to narrow the range, **thus making pointer $j$ closer to the elements less than `target`**.
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After the loop, $i$ points to the leftmost `target`, and $j$ points to the first element less than `target`, **therefore index $i$ is the insertion point**.
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=== "<1>"
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=== "<2>"
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=== "<3>"
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=== "<4>"
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=== "<5>"
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=== "<6>"
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=== "<7>"
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=== "<8>"
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Observe the following code. The operations in the branches `nums[m] > target` and `nums[m] == target` are the same, so these two branches can be merged.
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Even so, we can still keep the conditions expanded, as it makes the logic clearer and improves readability.
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```src
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[file]{binary_search_insertion}-[class]{}-[func]{binary_search_insertion}
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```
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!!! tip
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The code in this section uses "closed interval". If you are interested in "left-closed,right-open", try to implement the code on your own.
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In summary, binary search essentially involves setting search targets for pointers $i$ and $j$, which might be a specific element (like `target`) or a range of elements (like elements less than `target`).
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In the continuous loop of binary search, pointers $i$ and $j$ gradually approach the predefined target. In the end, they either find the answer or stop after crossing the boundary.
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